Problem: Divide the following rational expressions and simplify the result. $\dfrac{2x^2-4xy}{3x+6y}\div\dfrac{xy+y^2}{x^2+3xy+2y^2}=$
Solution: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator of the dividend $2x^2-4xy$ can be factored as $2x(x-2y)$ by factoring out $2x$. The denominator of the dividend $3x+6y$ can be factored as $3(x+2y)$ by factoring out a $3$. The numerator of the divisor $xy+y^2$ can be factored as $y(x+y)$ by factoring out a $y$. The denominator of the divisor $x^2+3xy+2y^2$ can be factored as $(x+y)(x+2y)$ using the sum-product pattern. Now the division looks as follows: $ \dfrac{2x(x-2y)}{3(x+2y)}\div\dfrac{y(x+y)}{(x+y)(x+2y)}$ To divide two rational expressions, we flip the divisor, multiply across, and simplify: [What's that?] $\phantom{=}\dfrac{2x(x-2y)}{3(x+2y)}\div\dfrac{y(x+y)}{(x+y)(x+2y)}$ $\begin{aligned}&=\dfrac{2x(x-2y)}{3(x+2y)}\cdot\dfrac{(x+y)(x+2y)}{y(x+y)}&\text{Flip the divisor.}\\\\\\\\ &=\dfrac{2x(x-2y)\cdot(x+y)(x+2y)}{3(x+2y)\cdot y(x+y)}&\text{Multiply across.}\\\\\\\\ &=\dfrac{2x(x-2y)\cancel{{(x+y)}}\cancel{{(x+2y)}}}{3y\cancel{{(x+2y)}}\cancel{{(x+y)}}}&\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{2x(x-2y)}{3y}\end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{2x(x-2y)}{3y}$, which is equivalent to $\dfrac{2x^2-4xy}{3y}$.